Question: A certain circle can be represented by the following equation. $x^2+y^2+12x+4y+15=0$ What is the center of this circle ? $($
The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2+y^2+12x+4y+15&=0\\\\ x^2+y^2+12x+4y&=-15\\\\ (x^2+12x)+(y^2+4y)&=-15 \text{(rearrange terms)}\\\\ (x^2+12x{+36})+(y^2+4y{+4})&=-15{+36}{+4}\end{aligned}$ Notice that we must add ${36}$ and ${4}$ on the right side of the equation, since we added them to the left side of the equation. [How did we get 36 and 4?] Writing the equation in standard form $\begin{aligned}(x^2+12x{+36})+(y^2+4y{+4})&=-15{+36}{+4}\\\\ (x+6)^2+(y+2)^2&=25\\\\ (x-(-6))^2+(y-(-2))^2&=5^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(-6,-2)$ and has a radius of $5$ units. Summary The circle is centered at $(-6,-2)$. The circle has a radius of $5$ units.